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Question about Maclaurin series - calculus

  1. Apr 27, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the Maclaurin series of the function https://webwork.wustl.edu/webwork2_files/tmp/equations/87/63afd4b6f3566e2a90aa420dc5d1821.png [Broken]
    c_3 =
    c_4 =
    c_5 =
    c_6 =
    c_7 =

    2. Relevant equations
    upload_2015-4-27_0-43-19.png

    3. The attempt at a solution
    (8x^2)[(9x) - (9x)^3/3! + (9x)^5/5! - (9x)^7/7! + ...]

    I got c_3 = 72, which is correct, and c_5 = (8x^2)[(9x^3)/3!] = 5832/3! (coefficient)

    Apparently that's wrong and it's supposed to be -116640/(5!).

    What did I do wrong? Thanks!
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Apr 27, 2015 #2

    PeroK

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    Are you not just muddling up which coefficient is which? If you've got 1/3! and the answer has 1/5! then those are different coefficients.
     
  4. Apr 27, 2015 #3

    Delta²

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    Isnt c4 supposed to be the coefficient of x^4 which is zero in this series?
     
  5. Apr 27, 2015 #4
    It's supposed to be c_5, not c_4, sorry.
     
    Last edited by a moderator: May 7, 2017
  6. Apr 27, 2015 #5

    PeroK

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    First, your coefficient should have a minus.

    Second, why do you think your answer and the answer given are different? Just because they look different doesn't mean they are!
     
  7. Apr 27, 2015 #6

    Delta²

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    Well PeroK is right, doing the calculations it turns out that 5832/3!=116640/5!. However i wonder if your teacher wanted you to do this assignment by manually calculating up to the 7th derivative of 8x^2sin(9x) and evaluating the derivatives at 0.
     
  8. Apr 27, 2015 #7

    PeroK

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    I suspect that since we have the coefficient of ##x^5##, whoever gave the answer preferred to have ##5!## in the denominator.
     
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